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3-Phenylpropene on reaction with $\mathrm{HBr}$ gives (as a major product)
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The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_2 \mathrm{CH}_3$
According to Markownikoff's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.
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